Double Pointers

[Extremist]

Hi there

I program in Linux, C++.

If you declare a char pointer in C++ , you would say

char *pointer;

And if you want to allocate memory for that pointer you would say

pointer = new char[50];

Now, if I want to declare a double pointer

char **pointer;

How would you allocate memory for such a double pointer?

[Silent1Mezzo]

Hi there

I program in Linux, C++.

If you declare a char pointer in C++ , you would say

char *pointer;

And if you want to allocate memory for that pointer you would say

pointer = new char[50];

Now, if I want to declare a double pointer

char **pointer;

How would you allocate memory for such a double pointer?

[CODE=cpp]char **pointer;
pointer= new *char[50];[/CODE]
That's how you can declare a double pointer

PS...I'm a c programmer...no t having to malloc it is really nice.

[Extremist]

[CODE=cpp]char **pointer;
pointer= new *char[50];[/CODE]
That's how you can declare a double pointer

PS...I'm a c programmer...no t having to malloc it is really nice.

Ok...
I did that

I get a segmentation fault

I have established that if only saying

char *pointer[50];

works nicely, but I was just wondering
Thanx for your input though.

PS I'm also used to C myself

[ilikepython]

Ok...
I did that

I get a segmentation fault

I have established that if only saying

char *pointer[50];

works nicely, but I was just wondering
Thanx for your input though.

PS I'm also used to C myself

I think double pointers are declared like this:
[code=cpp]
char **pointer;
pointer = new char*[50];
[/code]
I tried it and it works.

[emaghero]

Hi there

I program in Linux, C++.

If you declare a char pointer in C++ , you would say

char *pointer;

And if you want to allocate memory for that pointer you would say

pointer = new char[50];

Now, if I want to declare a double pointer

char **pointer;

How would you allocate memory for such a double pointer?

If by "double pointer" you mean a pointer to a pointer use the following:

char **p=new(char *[size]);//size MUST be of type int
//round brackets are for clarity
//the code will work without them

This will work for pointers of any type (Naturally you can't mix types).

Dynamic memory allocation is second nature to me.
All hail the new operator.

[Extremist]

Oh, thank you soooo much!!!!
You really helped me!!! I'm really greatful!
I've been struggling somehow with this, so thank you

Zel ;)

[Arvind07]

Kindly check this program for double pointer allocation
[code=c]
#include <stdio.h>

int main ()
{
int **p; /* 3*4 allocation */
int i,j;
p = malloc (3*sizeof (int));
printf ("Address of p:%p\n", p);

for (i=0; i<4; i++)
{
*(p+i) = malloc (4*sizeof (int));
printf ("Address (%p)\n", *(p+i));

}
printf ("Address of *p:%p\n", *p);
**p = 12;
*(*p+1) = 14;
*(*p+2) = 16;
*(*p+3) = 18;

**(p+1) = 5;
*(*(p+1)+1) = 6;
*(*(p+1)+2) = 7;
*(*(p+1)+3) = 8;

**(p+2) = 9;
*(*(p+2)+1) = 10;
*(*(p+2)+2) = 11;
*(*(p+2)+3) = 12;



for (i=0; i<3; i++)
{
for (j=0; j<4; j++)
{
printf (" Array (%d) \n", *(*(p+i)+j));
}
printf ("One row finished \n");
}
} // main ()
[/code]

[Silent1Mezzo]

Kindly check this program for double pointer allocation
#include <stdio.h>

int main ()
{
int **p; /* 3*4 allocation */
int i,j;
p = malloc (3*sizeof (int));
printf ("Address of p:%p\n", p);

for (i=0; i<4; i++)
{
*(p+i) = malloc (4*sizeof (int));
printf ("Address (%p)\n", *(p+i));

}
printf ("Address of *p:%p\n", *p);
**p = 12;
*(*p+1) = 14;
*(*p+2) = 16;
*(*p+3) = 18;

**(p+1) = 5;
*(*(p+1)+1) = 6;
*(*(p+1)+2) = 7;
*(*(p+1)+3) = 8;

**(p+2) = 9;
*(*(p+2)+1) = 10;
*(*(p+2)+2) = 11;
*(*(p+2)+3) = 12;



for (i=0; i<3; i++)
{
for (j=0; j<4; j++)
{
printf (" Array (%d) \n", *(*(p+i)+j));
}
printf ("One row finished \n");
}
} // main ()

You're kind of hijacking this post. You should start your own probably.

[j3n0vacHiLd]

2D array using a pointer to a pointer.

char **p;
p = new char*[50]; // p = new *char[50]; will give you a syntax error
*p = new char[50];
p[0][0] = 'a';
p[0][1] = 'b';
p[0][2] = 'b';

Just don't forget if you don't want any memory leaks.

delete [] *p;
delete [] p;

----

[evillestat]

2D array using a pointer to a pointer.

char **p;
p = new char*[50]; // p = new *char[50]; will give you a syntax error
*p = new char[50];
p[0][0] = 'a';
p[0][1] = 'b';
p[0][2] = 'b';

Just don't forget if you don't want any memory leaks.

delete [] *p;
delete [] p;

----

Not to dig up terribly old code, but I saw this and was curious.

I have not tried this, but does this work?


p = new char*[50]; // p = new *char[50]; will give you a syntax error
*p = new char[50];


As in, this would be the same as:

char p[50][50];

I was always told you had to use a for loop to allocate the memory dynamically for the double pointer 2-d array.

Thusly:

p = new int *[r];
for(i = 0; i < r; i++)
{
p[r] = new int[c];
}

Where R and C are rows and columns, respectively.

Any thoughts?

[Ganon11]

[CODE=cpp]p = new char*[50]; // p = new *char[50]; will give you a syntax error
*p = new char[50];
[/CODE]

means that p is a pointer to an array of 50 pointers to characters. The first pointer in this array, p[0] (or *p), points to an array of 50 characters. None of the other pointers in p point to anything.

[evillestat]

[CODE=cpp]p = new char*[50]; // p = new *char[50]; will give you a syntax error
*p = new char[50];
[/CODE]

means that p is a pointer to an array of 50 pointers to characters. The first pointer in this array, p[0] (or *p), points to an array of 50 characters. None of the other pointers in p point to anything.

So I was correct in thinking this would not work? Becuase if p[0] is pointing to an array of 50 characters, and nothing else is point anywhere, then it isn't a 2-d array of p[50][50] but p[0] = p[50]

So then you would have to set it another way, correct?

[Ganon11]

Exactly .